f(x)= array|c|c| x^3 ((1)/(x)), & x 0 0 & ,x=0 array

Question

$f(x)= \begin{array}{|c|c|} {x}^{3}\sin (\frac{1}{x}), & x
e 0 \ 0 & ,x=0 \end{array}$

Accepted Answer

Given the function:

$f(x)= \begin{gathered} {x}^{3}\sin (\frac{1}{x}), x
e 0 \ 0, x=0 \end{gathered}$

we need to find its second derivative at specific points.

First, let's compute the first derivative ${f}^{'}(x)$:

${f}^{'}(x)=3{x}^{2}\sin (\frac{1}{x})-x\cos (\frac{1}{x})$

Next, the second derivative ${f}^{''}(x)$ is:

${f}^{''}(x)=6x\sin (\frac{1}{x})-3x\cos (\frac{1}{x})-\cos (\frac{1}{x})-\frac{1}{x}\sin (\frac{1}{x})$

Therefore, evaluating the second derivative at $x=\frac{2}{\pi }$:

${f}^{''}(\frac{2}{\pi })=6(\frac{2}{\pi })\sin (\frac{\pi }{2})-3(\frac{2}{\pi })\cos (\frac{\pi }{2})-\cos (\frac{\pi }{2})-\frac{\pi }{2}\sin (\frac{\pi }{2})$

Since $\sin (\frac{\pi }{2})=1$ and $\cos (\frac{\pi }{2})=0$, this simplifies to:

${f}^{''}(\frac{2}{\pi })=\frac{12}{\pi }-\frac{\pi }{2}=\frac{24-{\pi }^{2}}{2\pi }$

Finally, note that ${f}^{'}(0)$ is not defined, as it involves terms like $\frac{1}{x}$ when $x=0$.

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